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Thread: Maximize volume of Cylinder

  1. #1
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    Maximize volume of Cylinder

    We want to build a cylindrical can with total surface area of 100 square inches. This surface area includes the top and bottom of the can. Find the dimensions of the can of this type that maximizes the volume of the can.


    I'm not getting the right answer... apparently the height should be twice the radius (according to the answer key) but...

    here's what i got:


    V = pi * r^2 * h
    SA : 2 * pi * r * h + 2*pi*r^2 = 100

    V is my optimizing function so i want it in terms of one variable right?

    Therefore, i can solve for h with SA and get that h = (50-pi *r^2)/ (pi*r)

    When i plug in this into my Volume function, i get v = r(50-pi*r^2)

    the derivative of v is just v'= 50-3r^2*pi

    i solve for r and get r = sqrt(50/(3pi))

    since, r = sqrt(50/(3pi)), h = 100/(3pisqrt(50/3pi)) apparently, this isn't the answer ;/
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  2. #2
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    Re: Maximize volume of Cylinder

    These are the right numbers, they probably just want it simplified.

    $r = \dfrac{5\sqrt{2}}{3\pi}$

    and similar for $h$
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  3. #3
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    Re: Maximize volume of Cylinder

    On the answer key, it didn't provide the actual solution but it just tells us that height is twice the radius h=2r
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  4. #4
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    Re: Maximize volume of Cylinder

    Quote Originally Posted by lc99 View Post
    On the answer key, it didn't provide the actual solution but it just tells us that height is twice the radius h=2r
    Well that's garbage. You were given a numeric surface area to meet.

    If you look at your answer you'll note that $h=2r$
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